Which bonds are formed by a carbon atom with \[s{{p}^{2}}\] - hybridisation?
A.$4\pi $ - bonds
B.$4\pi $- bonds and $2\sigma $- bonds
C.$1\pi $ - bond and $3\sigma $ - bonds
D.$4\sigma $- bonds

Hint:We are asked about which bonds are formed by carbon atoms with \[s{{p}^{2}}\] hybridisation. We will see here sigma and pi bonds are formed. Also we will use a formula.
Hybridisation = Number of bond pair + number of lone pair

Complete answer:
When the atomic orbitals overlap along the bond axis then sigma bonds are formed.
On the other hand when atomic orbitals overlap side by side then pi bonds are formed.
For hybridisation only sigma bonds are required, pi bonds do not have any role in hybridisation. So in the formula used above the number of bonds refers only to sigma bonds.
In the formula we also consider lone pairs. But we don’t have any options with lone pairs so we will not consider that in this question.
If the summation comes as 2, 3, 4, 5 then the hybridisation is $sp,\,s{{p}^{2}},\,s{{p}^{3}},\,s{{p}^{3}}d$ respectively.
In the question we are given with\[s{{p}^{2}}\] hybridisation. So from this we get to know that the hybridisation is 3. As lone pairs are not there and pi bonds do not contribute to hybridisation so we will find out the option with 3 sigma bonds.
In the first option we have $4\pi $- bonds, in the second option we have $4\pi $- bonds and $2\sigma $ - bonds, in the third option we have $1\pi $ - bond and $3\sigma $ - bond, in the fourth option we have $4\sigma $ - bonds.

So, the correct option is C.

Note:
The formula should be known. Also it should be noted that the pi bonds do not have a contribution in hybridisation.
If the hybridisation is 6 it will not be $s{{p}^{5}}$ as this is not possible. It will be $s{{p}^{3}}{{d}^{2}}$.